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Homework Help: Spherical Shell in External Field

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data
    An uncharged hollow conducting spherical shell with inner and outer radii a,b respectively
    is placed into otherwise uniform electric field E. Calculate the induced charge density at a and b and electric field everywhere
    Solution for E ,r>b
    [tex]E = E_0 (\widehat{E_0}+\frac{3\widehat{r}Cos(\theta)-\widehat{E_0}}{r^3}b^3)[/tex]


    2. Relevant equations
    Poisson Equation
    maxwells equation


    3. The attempt at a solution
    Obviously E = 0 for r<b
    The problem seems very similar to shell at radius R=b
    in which case using the
    [tex]V = \Sigma (A_l r^l + \frac{B_l}{r^{l+1}})P_l (Cos(\theta))[/tex]
    with boundary condition that V=0 in conductor gives the
    [tex]\sigma = \frac{3}{4\pi}E_0 Cos(\theta)[/tex] in outer surface (right answer)
    but wrong answer for the potential
    [tex]V = -E_0 Cos(\theta) (r-\frac{b^3}{r^2})[/tex]
    for the potential which is not the same as book when taking negative
    divergence
     
  2. jcsd
  3. Jun 28, 2010 #2

    kuruman

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    Is there a question here? The solution to this problem is identical to the ungrounded conducting shell of radius R = b in a (previously) uniform electric field. The Uniqueness Theorem guarantees that. Grounding makes no difference. Since the potential cannot be zero at infinity (why?), you can specify it to be zero (or any value you wish) on the surface.
     
  4. Jun 28, 2010 #3
    I get a different E field than the one in the solution which I give the problem statement. Im not sure if im doing it right and the book has a wrong solution. As seen in my attempt I assumed it was the same as a shell of radius R=b and proceeded to solve for the boundary conditions but I did not get the same answer as the book
     
  5. Jun 28, 2010 #4

    kuruman

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    Let's start with potentials. What is the book's answer for the potential? What is your answer for the potential? Write them down side by side and then we will figure out the rest.
     
  6. Jun 29, 2010 #5
    The book doesnt have a solution for the voltage just the E field, and my E field is
    [tex] E = E_0 Cos(\theta) (1+2\frac{b^3}{r^3} ) \widehat{r}-E_0 Sin(\theta) (1-\frac{b^3}{r^3} )\widehat{\theta} [/tex]
    with my E-field along z-axis.
    I get that from doing taking divergence of V and multiplying by -1.
     
  7. Jun 29, 2010 #6

    kuruman

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    Actually you need to take the negative gradient not divergence, but I think you did it right. However, you still did not tell me what your potential V(r) is. I need to see what it looks like, because that's the starting point.
     
  8. Jun 29, 2010 #7
    this is my potential
    from the boundary conditions at r=b and r infinity.
    [tex]
    V = -E_0 Cos(\theta) (r-\frac{b^3}{r^2})
    [/tex]
     
  9. Jun 29, 2010 #8

    kuruman

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    Good. This is the correct potential. You know it is correct because it is a solution to Laplace's equation and matches the boundary conditions. Now calculate the surface charge density from this using

    [tex]\sigma=-\frac{1}{\epsilon_0}\frac{\partial V}{\partial r}|_{r=b}[/tex]

    What do you get?
     
  10. Jul 1, 2010 #9
    [tex]
    \sigma = \frac{3}{4\pi}E_0 Cos(\theta)
    [/tex]
     
  11. Jul 1, 2010 #10

    kuruman

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    Can you show the details of your calculation? Specifically where did the factor of 4π in the denominator come from?
     
  12. Jul 2, 2010 #11
    No I took the derivative and thought translating SI to Gaussian made the 1/4pi but that is for k not epsilon.

    there is no 4 pi term but the solution in the book has a 4pi term which means none of my answers are the same as the book.
     
  13. Jul 3, 2010 #12

    kuruman

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    I somehow missed that you were working in gaussian units. First let me correct my expression in post #8 for anyone who cares to read this thread. In the SI system the correct expression for the surface charge density should be

    [tex]\sigma=-\epsilon_0\frac{\partial V}{\partial r}[/tex]

    In the gaussian system it should be

    [tex]\sigma=-\frac{1}{4\pi}\frac{\partial V}{\partial r}[/tex]

    Note that the potential, as you have written it, is the same in the gaussian and SI systems. So your expression for the surface charge density is correct after all. Your original question is that your potential is incorrect. What makes you think that?
     
    Last edited: Jul 3, 2010
  14. Jul 3, 2010 #13
    where does the [tex]4\pi
    [/tex]
    come from because I thought [tex]\epsilon_0=1
    [/tex].

    I think my potential is wrong because when I calculate the electric field I get
    [tex]
    E = E_0 Cos(\theta) (1+2\frac{b^3}{r^3} ) \widehat{r}-E_0 Sin(\theta) (1-\frac{b^3}{r^3} )\widehat{\theta}
    [/tex]

    but the book has this as the electric field
    [tex]
    E = E_0 (\widehat{E_0}+\frac{3\widehat{r}Cos(\theta)-\widehat{E_0}}{r^3}b^3)
    [/tex]

    where [tex]
    \widehat{E_0} = \widehat{z}
    [/tex]
     
  15. Jul 4, 2010 #14

    kuruman

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    The expression relating the surface charge density too the electric field is essentially derived from Gauss's Law. In the SI system Gauss's Law is

    [tex]\nabla \cdot E=\rho/\epsilon_0[/tex]

    In the gaussian system it is

    [tex]\nabla \cdot E=4 \pi \rho[/tex]

    To go from SI to gaussian you replace 1/ε0 with 4π (or ε0 with 1/4π).

    Have you tried reconciling the two expressions? What if you expressed theta-hat in your expression in terms of E0-hat and r-hat?
     
  16. Jul 4, 2010 #15
    [tex] \widehat{\theta} = \frac{\widehat{\rho} - \widehat{z}}{\sqrt{2}}[/tex] and
    [tex]\widehat{r} = \frac{\widehat{\rho} + \widehat{z}}{\sqrt{2}} [/tex]
    this gives
    [tex] \widehat{\theta} = \widehat{r} - \frac{\widehat{E_0}}{\sqrt{2}} [/tex]
    which gives

    [tex] E = E_0 \frac{\sin(\theta)}{\sqrt{2}} (1-\frac{b^3}{r^3})\widehat{E_0} + [(1+2\frac{b^3}{r^3})\cos(\theta)-(1-\frac{b^3}{r^3})\sin(\theta)]\widehat{r} [/tex]
    this is only the same if on the left term sine is equal to 1 and in the right term the sine is a cosine
     
    Last edited: Jul 4, 2010
  17. Jul 4, 2010 #16

    kuruman

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    These expressions are not generally true; they are valid when θ = 45o
    This is what you do:
    1. Start from your expression and add and subtract

    [tex]E_0\frac{r^3}{b^3}Cos(\theta)\widehat{r}[/tex]

    Note that what you get already has part of what you need
    [tex]3 E_0 Cos(\theta)\frac{r^3}{b^3}\widehat{r}[/tex]

    2. For the rest of the terms factor out
    [tex]\left (\widehat{r}Cos(\theta)-\widehat{\theta}Sin(\theta) \right)[/tex]

    which is really [tex]\widehat{z}[/tex] and you are done.
     
    Last edited: Jul 4, 2010
  18. Jul 5, 2010 #17
    that sounds right.
    Is this the most concise way to solve this problem.
     
  19. Jul 6, 2010 #18

    kuruman

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    Probably. Note that the answer in the book is given in a mixed bag of unit vectors, cartesian and spherical. The quickest way is to start from the potential in spherical coordinates to match the symmetry of the problem and then take the gradient. After that, you need to get rid of theta-hat. You might be able to find a clever way of taking the gradient using the chain rule so that the E-field in the book immediately pops out, but I am not sure if it is more concise because you still need an extra derivation step at least.
     
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